To write the net ionic equation for the reaction \(\text{SnCl}_2(aq) + 2\text{KOH}(aq) \rightarrow \text{Sn(OH)}_2(s) + 2\text{KCl}(aq)\), follow these steps:
1. **Write the complete ionic equation** by breaking all the aqueous compounds into their ions:
\[\text{Sn}^{2+}(aq) + 2\text{Cl}^-(aq) + 2\text{K}^+(aq) + 2\text{OH}^-(aq) \rightarrow \text{Sn(OH)}_2(s) + 2\text{K}^+(aq) + 2\text{Cl}^-(aq)\]
2. **Identify and remove the spectator ions**, which are ions that appear on both sides of the equation and do not participate in the actual reaction:
The spectator ions in this case are \(2\text{K}^+(aq)\) and \(2\text{Cl}^-(aq)\).
3. **Write the net ionic equation** by omitting the spectator ions:
\[\text{Sn}^{2+}(aq) + 2\text{OH}^-(aq) \rightarrow \text{Sn(OH)}_2(s)\]
This is the net ionic equation for the given reaction. It shows the actual chemical species that participate in the reaction:
\[\text{Sn}^{2+}(aq) + 2\text{OH}^-(aq) \rightarrow \text{Sn(OH)}_2(s)\]
The following chemical reaction takes place in aqueous solution:
SnCl_2(aq)+2KOH(aq)->Sn(OH)_2(s)+2KCl(aq)
Write the net ionic equation for this reaction
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